Respuesta :
Answer:
The reaction will be non spontaneous at these concentrations.
Explanation:
[tex]AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)[/tex]
Expression for an equilibrium constant [tex]K_c[/tex]:
[tex]K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-][/tex]
Solubility product of the reaction:
[tex]K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13} [/tex]
Reaction between Gibb's free energy and equilibrium constant if given as:
[tex]\Delta G^o=-2.303\times R\times T\times \log K_c[/tex]
[tex]\Delta G^o=-2.303\times R\times T\times \log K_{sp}[/tex]
[tex]\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}][/tex]
[tex]\Delta G^o=69,117.84 J/mol=69.117 kJ/mol[/tex]
Gibb's free energy when concentration [tex][Ag^+] = 1.0\times 10^{-2} M[/tex] and [tex][Br^-] = 1.0\times 10^{-3} M[/tex]
Reaction quotient of an equilibrium = Q
[tex]Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}[/tex]
[tex]\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)[/tex]
[tex]\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])[/tex]
[tex]\Delta G=40.588 kJ/mol[/tex]
- For reaction to spontaneous reaction: [tex]\Delta G<0[/tex].
- For reaction to non spontaneous reaction: [tex]\Delta G>0[/tex].
Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations
